Question

two parallel sides of an isosceles trapezium are 10 cm and 20 cm and it's non parallel sides are each equal to 13 cm. find the area of trapezium ​

Answer 1

Answer:

Area of the trapezium is 180 cm².

Step-by-step explanation:

The figure is drawn below,

From the figure, it is clear that the lengths CD and EF are each equal to 5 cm due to symmetry of the figure.

Consider Δ BCD which is a right angled triangle.

Using Pythagoras Theorem,

\begin{gathered}BC^2=CD^2+BD^2\\13^2=5^2+BD^2\\169=25+BD^2\\BD^2=169-25\\BD^2=144\\BD=\sqrt{144}=12\ cm\end{gathered}

BC

2

=CD

2

+BD

2

13

2

=5

2

+BD

2

169=25+BD

2

BD

2

=169−25

BD

2

=144

BD=

144

=12 cm

Now, area of trapezium ABCF is given as:

\begin{gathered}Area=\frac{1}{2}\times (\textrm{Sum of parallel sides}\times \textrm{Height}\\Area=\frac{1}{2}\times (10+20)\times 12\\Area=\frac{1}{2}\times 30\times 12=15\times 12=180\ cm^2\end{gathered}

Area=

2

1

×(Sum of parallel sides×Height

Area=

2

1

×(10+20)×12

Area=

2

1

×30×12=15×12=180 cm

2

Therefore, the area of the trapezium with parallel sides 10 cm and 20 cm is 180 cm².

Step-by-step explanation:

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